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Mathematics 9 Online
OpenStudy (adi3):

Will Medal Please help!! Use Quadratic formula to solve for x: (x+2)(x-1) = 5

OpenStudy (anonymous):

Start off by putting into \[ax^2+bx+c=0\] form

OpenStudy (adi3):

ok, wait let me solve it first

OpenStudy (anonymous):

Take your time

OpenStudy (adi3):

ok, x^2 +1x -7

OpenStudy (anonymous):

Good now we have the quadratic formula which is the following \[x = \frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] where a = 1, b= 1, and c = -7

OpenStudy (adi3):

ok, wait again sorry

OpenStudy (adi3):

Answer is 2,-1

OpenStudy (anonymous):

Are you sure?

OpenStudy (adi3):

yes

OpenStudy (adi3):

no no i am wrong wait

OpenStudy (adi3):

the answer is 2.15 and -3.15

OpenStudy (adi3):

i took the wrong equation

OpenStudy (anonymous):

\[x= \frac{ -1 \pm \sqrt{1^2-4(1)(-7)} }{ 2(1) } \implies x = \frac{ -1 \pm \sqrt{29} }{ 2 }\] correct?

OpenStudy (adi3):

yes

OpenStudy (anonymous):

That should \[x \approx -3.19,~~ \text{and}~~ x \approx 2.19\]

OpenStudy (adi3):

can i ask you an another quetion pls

OpenStudy (anonymous):

Sure thing

OpenStudy (adi3):

Use Quadratic formula to solve for x: \[x + \frac{ 1 }{ x+2 } = 4\]

OpenStudy (anonymous):

Same thing as earlier, try putting it into \[ax^2+bx+c=0\]

OpenStudy (adi3):

how will i do that

OpenStudy (anonymous):

You can multiply through by (x+2)

OpenStudy (adi3):

??

OpenStudy (anonymous):

I mean you can find a common denominator which in this case is (x+2) if you like

OpenStudy (adi3):

can we cross multiply

OpenStudy (anonymous):

You can do as following \[\frac{ 1 }{ x+2 }=4-x \] now you may do as you like

OpenStudy (anonymous):

\[\frac{ 1 }{ (x+2) } = (4-x)\]

OpenStudy (adi3):

ok

OpenStudy (adi3):

wait

OpenStudy (adi3):

is it = to 4x-x-5??

OpenStudy (anonymous):

Hmm, not quite, \[\frac{ 1 }{ (x+2) } = (4-x) \implies 1 = (4-x)(x+2)\]

OpenStudy (adi3):

yeah then we multiply (4-x)(x+2) = 4x - x +6

OpenStudy (adi3):

an we bring it to the other side

OpenStudy (adi3):

-4x + x -6 + 1

OpenStudy (adi3):

= -4x + x -5 = 0

OpenStudy (adi3):

right??

OpenStudy (anonymous):

Remember, we want our equation in \[ax^2+bx+c=0\] form, so what you wrote, does that make sense? Think of what you did in the previous problem

OpenStudy (adi3):

x^2-4x-5

OpenStudy (adi3):

= 0

OpenStudy (anonymous):

\[(4-x)(x+2)=1 \implies 4x+8-x^2-2x =1 \implies 2x+8-x^2=1 \] so we can set this up as \[x^2-2x-7=0\]

OpenStudy (adi3):

ok. sorry i was offline, i had a lag

OpenStudy (adi3):

answer is = 1.9, -0.9 right??

OpenStudy (adi3):

@iambatman

OpenStudy (anonymous):

Can you show how you got that?

OpenStudy (adi3):

ok |dw:1444558479751:dw|

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